A ball is spun with angular acceleration $$\alpha = 6t^2 - 2t$$ where $$t$$ is in second and $$\alpha$$ is in rad s$$^{-2}$$. At $$t = 0$$, the ball has angular velocity of 10 rad s$$^{-1}$$ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is

We are given angular acceleration $$\alpha = 6t^2 - 2t$$ rad s$$^{-2}$$, with initial conditions $$\omega(0) = 10$$ rad s$$^{-1}$$ and $$\theta(0) = 4$$ rad.

Integrating the angular acceleration with respect to time gives $$\omega = \int \alpha \, dt = \int (6t^2 - 2t) \, dt = 2t^3 - t^2 + C_1$$. Imposing $$\omega(0) = 10$$ rad s$$^{-1}$$ yields $$C_1 = 10$$, so $$\omega = 2t^3 - t^2 + 10$$. Further integrating this angular velocity yields $$\theta = \int \omega \, dt = \int (2t^3 - t^2 + 10) \, dt = \frac{2t^4}{4} - \frac{t^3}{3} + 10t + C_2$$, hence $$\theta = \frac{t^4}{2} - \frac{t^3}{3} + 10t + C_2$$. Applying $$\theta(0) = 4$$ rad gives $$C_2 = 4$$, and thus $$\theta = \frac{t^4}{2} - \frac{t^3}{3} + 10t + 4$$. The correct answer is Option B.